设无穷非负数列${a_n}$满足$a_n+a_{n+2}ge2 a_{n+1},sumlimits_{i=1}^{n}{a_i}le1$,证明:
$0le a_n-a_{n+1}ledfrac{2}{n(n+1)}$
证明：由题意$0le a_nlesumlimits_{i=1}^{n}{a_i}le1$又由于$1ge a_{m}-a_nge(m-n)(a_{n+1}-a_n)$
(凸数列性质,有定义易得)
故当$m>n$时,$a_{n+1}-a_nle dfrac{1}{m-n}longrightarrow 0 (mlongrightarrow infty)$
右边由$a_{i}ge a_n+(i-n)(a_{n+1}-a_n)=a_{n+1}+(n+1-i)(a_{n}-a_{n+1})ge(n+1-i)(a_{n}-a_{n+1})$
故$1gesumlimits_{i=1}^{n}{a_i}gesumlimits_{i=1}^{n}{(n+1-i)(a_{n}-a_{n+1})}=dfrac{1}{2}n(n+1)(a_n-a_{n+1})$
即得$0le a_n-a_{n+1}le<dfrac{2}{n(n+1)}$
注:此类数列特例$a_n=dfrac{6}{pi n^2}$